What is the equation of the normal line of #f(x)=x^3-11x^2-5x-2# at #x=0#?
To find the slope of the normal line, first find the slope of the tangent line. Since the tangent line and normal line are perpendicular, their slopes will be the opposite reciprocals of one another.
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The equation of the normal line of f(x)=x^3-11x^2-5x-2 at x=0 is y = -2.
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The equation of the normal line to the curve of a function at a given point is given by the negative reciprocal of the derivative of the function at that point.
First, find the derivative of (f(x) = x^3 - 11x^2 - 5x - 2): [f'(x) = 3x^2 - 22x - 5]
At (x = 0), the slope of the tangent line is (f'(0) = -5).
The slope of the normal line is the negative reciprocal of the slope of the tangent line, so the slope of the normal line is (1/5).
Therefore, the equation of the normal line at (x = 0) is (y = (1/5)x + b).
To find (b), substitute (x = 0) and (y = f(0)) into the equation: [f(0) = 0^3 - 11(0)^2 - 5(0) - 2 = -2] So, (b = -2).
Therefore, the equation of the normal line at (x = 0) is (y = (1/5)x - 2).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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